Problem: A permutation p is an ordered group of numbers p1, p2, ..., pn, consisting of n distinct positive integers, each is no more than n. We'll define number n as the length of permutation p1, p2, ..., pn.

Simon has a positive integer n and a non-negative integer k, such that 2k ≤ n. Help him find permutation a of length 2n, such that it meets this equation: <image>.

Input

The first line contains two integers n and k (1 ≤ n ≤ 50000, 0 ≤ 2k ≤ n).

Output

Print 2n integers a1, a2, ..., a2n — the required permutation a. It is guaranteed that the solution exists. If there are multiple solutions, you can print any of them.

Examples

Input

1 0


Output

1 2

Input

2 1


Output

3 2 1 4


Input

4 0


Output

2 7 4 6 1 3 5 8

Note

Record |x| represents the absolute value of number x. 

In the first sample |1 - 2| - |1 - 2| = 0.

In the second sample |3 - 2| + |1 - 4| - |3 - 2 + 1 - 4| = 1 + 3 - 2 = 2.

In the third sample |2 - 7| + |4 - 6| + |1 - 3| + |5 - 8| - |2 - 7 + 4 - 6 + 1 - 3 + 5 - 8| = 12 - 12 = 0.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
string to_string(string s) { return '"' + s + '"'; }
string to_string(char s) { return string(1, s); }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A>
string to_string(A);
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool f = 1;
  string r = "{";
  for (const auto& x : v) {
    if (!f) r += ", ";
    f = 0;
    r += to_string(x);
  }
  return r + "}";
}
void debug_out() { cout << "\n"; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cout << " " << to_string(H);
  debug_out(T...);
}
const long long N = 100043;
long long fact[N];
long long add(long long x, long long y) {
  x += y;
  while (x >= 1000000007) x -= 1000000007;
  while (x < 0) x += 1000000007;
  return x;
}
long long mul(long long x, long long y) { return (x * 1ll * y) % 1000000007; }
long long mulmod(long long a, long long b) {
  long long res = 0;
  a = a % 1000000007;
  while (b > 0) {
    if (b % 2 == 1) res = (res + a) % 1000000007;
    a = (a * 2) % 1000000007;
    b /= 2;
  }
  return res % 1000000007;
}
long long binpow(long long x, long long y) {
  long long z = 1;
  while (y > 0) {
    if (y % 2 == 1) z = mul(z, x);
    x = mul(x, x);
    y /= 2;
  }
  return z;
}
long long inv(long long x) { return binpow(x, 1000000007 - 2); }
long long divide(long long x, long long y) { return mul(x, inv(y)); }
void precalc() {
  fact[0] = 1;
  long long i;
  for (i = 1; i < N; ++i) fact[i] = mul(i, fact[i - 1]);
}
long long C(long long n, long long k) {
  if (k > n) return 0;
  return divide(fact[n], mul(fact[n - k], fact[k]));
}
void solve() {
  int n, k, i, t;
  cin >> n >> k;
  vector<int> a, b;
  for (i = 1; i < 2 * n + 1; ++i)
    if (i % 2 == 0)
      a.emplace_back(i);
    else
      b.emplace_back(i);
  sort((a).rbegin(), (a).rend());
  sort((b).rbegin(), (b).rend());
  i = 0;
  while (i < k) {
    t = a[2 * i];
    a[2 * i] = a[2 * i + 1];
    a[2 * i + 1] = t;
    i++;
  }
  for (i = 0; i < n; ++i) cout << a[i] << " " << b[i] << " ";
  cout << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}